Principle Current source

Discussion in 'General Electronics Chat' started by phuzionz, Jan 29, 2009.

  1. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    Dear members,

    http://www.pi4hmd.nl/menu/varicon/v4150/v43/artikel07.htm
    Schematicc 1:
    Can somebody explain me how these current source is working? Why is the ground symbol connected to the collector?
    The zener sets always 5.7V on the base.
    The emitter resistor controls the Ic current, in other words, it stays always the same.
    Is that right?
    Why is that circuit grounded on the collector?
     
    Last edited: Jan 30, 2009
  2. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
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    It did take a while to figure this one out! It looked like the polarity of the transistor was wrong. However, if you look at the current source, they've grounded the positive end...so it will work.

    Positive ground circuits are used in a lot of situations where it might be awkward to interface it to an outside source or signal. For example in telephone office equipment, everything has a positive ground. To interface telephone equipment to negative ground circuits can be really tricky. (I know, I've had to do it numerous times!

    Anyway..the circuit functions identically, regardless of the ground point.

    eric

    Hope this helps.
     
  3. hgmjr

    Moderator

    Jan 28, 2005
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    I believe your question is concerning the NPN which has its collector referenced to ground. You will note that the 50V power supply has its positive terminal referenced to ground as well. This preserves the requirement that the emitter in an NPN be more negative than the potential at its collector.

    hgmjr
     
  4. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    But is that not a short then? The plus is fysically connected with the ground.
    Why they don't use my second circuit then? What is then the real difference between the two circuits.
     
    Last edited: Jan 29, 2009
  5. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    Can somebody help me with this. It's very important that ik know exactly what happened in the circuit.
    Why is the "ground" connected with the puls of the voltage supply, is that not a short?
     
  6. hgmjr

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    Jan 28, 2005
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    In your redrawn sketch there is no way to determine which is the emitter and which is the collector. This makes it impossible to comment on the validity of the circuit.

    hgmjr
     
  7. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    Sorry, i mean in the link that i gave in the first post. the colector is grounded, i don't understand that.
     
  8. hgmjr

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    Jan 28, 2005
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    In the case of the circuit in question, imagine if you will that the ground symbol is moved to the negative side of the battery instead of the positive side. That would not change the functioning of the circuit. It would change the values of voltages measured but the circuit would operate the same. Would the circuit be more digestible to you then?

    hgmjr
     
  9. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    Well yes i think so. the purpose of that ground signal is to make the voltages negative. How can i do that in practice?

    one more question, when i take Rc = 500ohm, is it normal that the transistor is not working well anymore .
     
  10. hgmjr

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    Jan 28, 2005
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    Since the collector current is 0.1 Amps then V=ir results in 50 volts. That means that the transistor is in saturation. That would prevent it from working.

    hgmjr
     
  11. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    Yes i agree with you, but normally when he is un saturation, you would measure a few volts, because on the graph (Ic versus Vce), it is close to the zero point.
    I build up the same circuit, only with other values. Vcc = 12V
    Zener= 5.6V, Rb= 670 ohm
    Re = 150 ohm, Rc= 500
    when i measure with these values, i measure for Vc 8V and for Ve more or less the same, is that not strange?
    Normally there have to flow the Ic(sat), but measured vaue is also less. Do you know why?
     
    Last edited: Jan 30, 2009
  12. hgmjr

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    Jan 28, 2005
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    A value of 8 volts for the voltage drop across Rc does sound a bit low. I would expect the conditions you have created using the new values would still result in a saturated state for the transistor.

    hgmjr
     
  13. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    8.8V to be precise. I will build up the circuit again.
    Thank you for your help
     
  14. hgmjr

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    Jan 28, 2005
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    On reflection, the value of 8.8V or so is probably not too far off. Remember that there will be close to a couple of volts dropped across the 150 Ohm emitter resistor.

    hgmjr
     
  15. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    I think that normal effect, I tried it now with 1000 ohm, and measure te current through Rc, but this is also less 8.4mA, thus that means that the voltage drop is also less.
     
  16. hgmjr

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    Jan 28, 2005
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    Keep in mind that the transistor is in saturation. That means that neglecting the 200 millivolts or so dropped across the transistor's emitter-collector means that the current is 12V/1150 Ohms. That comes to around 10 milliamps.

    hgmjr
     
  17. hgmjr

    Moderator

    Jan 28, 2005
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    You do realize that once a transistor enters saturation it no longer acts like a constant current source, right?

    hgmjr
     
  18. phuzionz

    Thread Starter Active Member

    Dec 5, 2008
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    Well yes that's right.

    I have checked that, and the currect was decreased!
     
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